In this screencast I am going to derive the geometric distribution, and I am going to
show an example of its use. Its a pretty simple distribution, and its a rather straightforward
derivation. This is one of the more simple of the distribution functions for discreet variables.
The geometric distribution is used to describe or predict the probability that you will have
a number of successful trials before having a failure, or having failures before having
a success. Lets do the latter where we're going to carry out trials until we have a
success. We might get lucky and the first trial might be a success, so then we would have
success on the first trial. OK, we might have situations where the first trial is a failure,
denoted by an f, and then we have a success. We might have two failures before we have
a success. We might have to go three failures before a success, so a success on the fourth
trial, and so on. Here we're going to denote x is going to be the number of trial, and
generally we are using the geometric distribution when we are asked "What is the probability
that the first success is on the xth trial?" Now if you had it reversed where you had,
you want to determine whats the probability that you have the first failure on the xth trial,
then you could adapt this, but you would just have to change around what you're using in
these equations. So in general we're going to have trials, and we're going to say we
have failures, success. So these dot, dot ,dots are all failures. So ultimately we are
going to have a success, and once we have the success then we're done. What is the probability
that the first success is on the xth trial. So this would be x, the number of trials.
This then, omitting the final success at the end, would be x minus one. To calculate the
probability of having x minus one failures that's just one minus p, where p is the probability
of success, raised to the x minus one, and if you watched my screencast on the binomial
distribution you noted that we had in the binomial distribution, we had like an n choose
x because each of the different independent combinations contribute to the overall probability. We
don't have that because how many ways are there to have x minus one failures into x
minus one spots, there is only one way to put these failures in there, so we don't need
to have anything out front here, and then whats the probability of having a success
at the end? That's just p. This will give us the probability that the first success
is in the xth trial. I have an example here. You work for a pharmaceutical company that
makes a product called "Curesall." What is the probability that the first failure occurs
on the 12th batch? And long term production has indicated that the probability of a batch
being successful is about 94.7 percent. So to solve this we can use the geometric distribution.
Now, you notice that up here in the geometric distribution, p is the probability of a success,
but we're going to call a success in this case a failure, so really we have to change
our p to be the probability of a failure, which in our terms is a success, because the
success in the geometric series is always is what occurs last, so we're going to go
success, success, success, success, failure. The probability of that happening is just
11 successes, and the probability of a success is 1 minus 0.053, or 0.947, so we're going
to have 11 successes, and then we're going to have a failure. So when we do the math,
plug this into our calculator, this ends up being about 3 percent. What that means in
conclusion there is a 3 percent chance that the first failure occurs on the 12th batch.